[Math] Proving $mn = 0$ implies $m = 0$ or $n=0$ for all $m, n \in \mathbb{N}$ using Peano Axioms

Question

I tried using induction on the set
$$S = \{ n \in \mathbb N : m \cdot n = 0 \implies m = 0 \vee n = 0, \forall m \in \mathbb N \}$$

1. Base step: $0 \in S$ because $m \cdot (0) = 0$ using the definition of multiplication of natural numbers. Then $n$ would be 0.
2. Inductive step: If $m \cdot n = 0 \implies m = 0 \vee n = 0$, then $m \cdot n^+ = 0 \implies m = 0 \vee n^+ = 0$.

Should I consider both cases when trying to prove the inductive step? I'm not even sure if that is what I should be proving or if the definition of $S$ is even right. Would a proof by contradiction be more appropriate?

Answer 1

Do it by contra-positive: suppose $m\neq 0\neq n$. There exist $k,l\in \Bbb N$ such that $m=k^+$ and $n=l^+$, (why?).

Thus $mn=ml^+=ml+m=ml+k^+=(ml+k)^+$.